I want to calculate the minimum surface area of a (closed) box for a given volume. So let’s say I have a given volume V (e.g. V=10m^3). And I need a box where all the surface area is as minimal as possible. This would be a great starting point if I knew how to calculate that.
To make things more complicated: Let’s say I have a given volume V but also I have a limit for the height of the box. So the height would be a certain measurement h (or higher), length and width would still be variable.
If you could answer my first question that would already be great. I think for the second problem I could also just do the same calculation/minimization and just work with a given area A (A = V/h).
$\endgroup$ 02 Answers
$\begingroup$Say that the Surface area is given by
$$A=2(ab+bc+ca).$$
Then, from the property that the Geometric Mean is always less that or equal to the Arithmetic Mean ($AM-GM$), we get
$$\frac{ab+bc+ca}{3}\geq\sqrt[3]{(abc)^2}.$$
Multiplying by $6$ gives
$$2(ab+bc+ca)\geq 6\sqrt[3]{(abc)^2},$$
where
$$abc=10m^3.$$
$\endgroup$ 3 $\begingroup$HINT
Indicating with $x,y,z$ the sides of the box we have
- $S=2(xy+yz+zx)$ surface to minimize
with the constraint
- $V=xyz=10$
